An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20.0°C.

(a) When placed in a room at a temperature of-4°C, will it gain time or losetime?

It will lose time.

It will gain time.

(b) How much time will it gain or lose every hour?

a) T ' = 0.999 s, b) t = 3596.4 s

Explanation:

The angular velocity of a simple pendulum is

w = √g / L

The angular velocity, frequency and period are related

w = 2π f = 2π / T

2π / T = √ g / L

T = 2π √ L / g

L = T² g / 4π²

L = 1² 9.8 / 4π²

L = 0.248 m

To know the effect of the temperature change let's use the thermal expansion ratios

ΔL = α L ΔT

ΔL = 24 10⁻⁶ 0.248 (-4 - 20)

ΔL = 142.8 10⁻⁶ m

Lf - L = - 142. 8 10⁻⁶

Lf = 142.8 10⁻⁶ + 0.248

Lf = 0.2479 m

Let's calculate new period

T ' = 2π √ L / g

T ' = 2π √ (0.2479 / 9.8)

T ' = 0.999 s

We can see that the value of the period is reduced so that the clock is delayed

b) change of time in 1 hour

When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is

t = 3600 0.999

t = 3596.4 s

Therefore the clock is delayed almost 4 s